MULTQ3 - Multiples of 3
There are N numbers a[0], a[1] ... a[N - 1]. Initially all are 0. You have to perform two types of operations :
- Increase the numbers between indices A and B (inclusive) by 1. This is represented by the command "0 A B"
- Answer how many numbers between indices A and B (inclusive) are divisible by 3. This is represented by the command "1 A B".
Input
The first line contains two integers, N and Q. Each of the next Q lines are either of the form "0 A B" or "1 A B" as mentioned above.
Output
Output 1 line for each of the queries of the form "1 A B" containing the required answer for the corresponding query.
Sample
Input: 4 7 1 0 3 0 1 2 0 1 3 1 0 0 0 0 3 1 3 3 1 0 3 Output: 4 1 0 2
Constraints
1 ≤ N ≤ 100000
1 ≤ Q ≤ 100000
0 ≤ A ≤ B ≤ N - 1
hide comments
jenishmonpara:
2020-05-19 19:04:35
AC in one go
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fighter_4:
2020-04-25 17:36:58
guys there is nothing tricky in 9th test case, you may have done any small mistake if you are getting WA,, took me 4 hours to debug,,,, but finally AC :D, my 20th. |
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perdedor_9:
2020-04-14 10:16:36
naive solution with optimization,AC in one go |
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aditya15:
2020-04-02 12:23:53
Test cases upto case 9 are simple sum queries without any updates, so even a wrong code may pass cases till 9. Just use Segment Trees + lazy propagation as mentioned below. No need to even use Fast IO |
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shivam_vc:
2020-03-31 10:03:49
got wa on 9th test case .can someone help ?
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sangmai:
2020-02-21 02:43:05
I constantly get WA in test case 9. Is there anyone with the same issue?
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subodh898:
2019-11-15 06:24:49
segment tree + Lazy +Fast I/O only |
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sourav1996:
2019-09-20 19:47:30
Can it be solved without lazy propagation? |
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rezagoodarzi:
2019-08-08 06:03:48
badihi jat |
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landofkings:
2019-07-28 05:31:57
A good problem to learn lazy propogation. |
Added by: | Varun Jalan |
Date: | 2010-09-12 |
Time limit: | 1s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: NODEJS OBJC VB.NET |
Resource: | own problem |